an alloy of metals is 20% copper. another alloy is 70% copper.how much of each should be used to make 80 gram of an alloy that is 50% copper?
10% copper 5% something else. I think
To determine the quantities of each alloy necessary to create an 80-gram mixture containing 50% copper, you can follow these steps:
Let's assume the first alloy contains x grams of copper and the second alloy contains y grams of copper.
1. Set up a system of equations based on the given information:
- For the copper content: 0.20x + 0.70y = 0.50(80)
(Since the target mixture should have 50% copper, and 80 grams in total)
- For the total weight: x + y = 80
2. Rearrange the second equation to solve for x:
x = 80 - y
3. Substitute the expression for x in the first equation:
0.20(80 - y) + 0.70y = 0.50(80)
4. Simplify and solve for y:
16 - 0.20y + 0.70y = 40
0.50y = 40 - 16
0.50y = 24
y = 24 / 0.50
y = 48
Therefore, the second alloy should contain 48 grams of copper.
5. Calculate the quantity of the first alloy (x):
x = 80 - y
x = 80 - 48
x = 32
Thus, the first alloy should contain 32 grams of copper.
To summarize, in order to obtain an 80-gram alloy containing 50% copper, you should use 32 grams of the first alloy (20% copper) and 48 grams of the second alloy (70% copper).